I am going through a fairly comprehensive sketch of a proof of the following theorem:
A positive integer $n \in \mathbb{Z}$ may be expressed as a sum of two squares $n = x^2 + y^2$ for $x,y \in \mathbb{Z} \iff ord_q(n)$ is even for all primes $q \equiv 3\ (\mod 4)$.
Note that $ord_q(n)$ is the largest integer m such that $q^m | n$ and $q^{m+1} \nmid n$.
The proof is this:
$n=a^2+b^2 \iff n=a^2+b^2=N(\alpha)$ where $\alpha = a+bi \in \mathbb{Z}[i]$ is a Gaussian integer. Such an integer $\alpha$ must be a product of norms of Gaussian primes which are: $2, p$ for any prime $p \equiv\ 1(\mod 4)$, and $q^2$ for any prime $q \equiv\ 3(\mod 4)$.
I know that $\mathbb{Z}[i]$ is a PID and thus a UFD and so $\alpha$ can be expressed as a product of irreducibles (which in $\mathbb{Z}[i]$ are the Gaussian primes) and so primes...
...but why is $\alpha$ then a product of norms of primes? Also, any insight as to why the primes are as stated: "...$2, p$ for any prime $p \equiv\ 1(\mod 4)$, and $q^2$ for any prime $q \equiv\ 3(\mod 4)$."
I think there's a typo in the quoted passage - it means to say that $n$ (the integer-integer, not the Gaussian integer $\alpha$) must be a product of norms, and this follows immediately from the multiplicativity of the norm: suppose $\alpha = g_1g_2\ldots g_n$ is a factorization of $\alpha$ into Gaussian primes; then $n=N(\alpha)=N(g_1)N(g_2)\ldots N(g_n)$ by that multiplicativity, so $n$ is (as stated) a product of norms of Gaussian primes. The last line is then talking about the possible norms: $2=|1+i|$, $p=|a+bi|$ (where $a^2+b^2=p$) for $p\equiv 1\pmod 4$, and $q^2=|q|$ for $q\equiv 3\pmod 4$.